PROOF 56
Prove.
a² + b² = c²
On the Euclidean plane.
A way to read this proof
First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.
Definition D1
In triangle ABC, let ∠C = 90° and set a = BC, b = CA, c = AB. ∠A, ∠B denote the interior angles at those vertices.
Proof.
(1)The two sides enclosing the right angle are a, b, and the side opposite the right angle is c.convention
(2)The side opposite ∠A is a and the side opposite ∠B is b.(1), convention
∎
Lemma L1
∠A + ∠B = 90°, and ∠A and ∠B are both acute.
Proof.
(1)∠A + ∠B + ∠C = 180°A2
(2)∠C = 90° and 180° = 90° + 90°D1, convention
(3)(∠A + ∠B) + 90° = 90° + 90°(1), (2), A20, E
(4)∠A + ∠B = 90°(3), A21
(5)Since ∠B > 0, ∠A < 90°, and since ∠A > 0, ∠B < 90°.(4), A10
∎
Lemma L2
Placing four triangles congruent to ABC in Arrangement I inside a square of side a + b gives (a+b)² = 2ab + c²
Proof.
(1)A square PQRS of side a + b exists.A3
(2)On sides PQ, QR, RS, SP take points E, F, G, H with PE = a, QF = a, RG = a, SH = a.(1), convention
(3)EQ = FR = GS = HP = b(1), (2), A15
(4)The angles of the square are right angles, so triangle PEH has a right angle at P with legs PE = a, PH = b.(1), (2), (3), convention
(5)Triangles EQF, FRG, GSH likewise have right angles at Q, R, S with legs a and b.(2), (3), (4)
(6)The four triangles agree with triangle ABC in two sides and the included angle, so all are congruent to ABC.(4), (5), D1, A5
(7)Corresponding sides are equal, so EH = FE = GF = HG = c(6), A5
(8)Corresponding angles are equal, so ∠PEH = ∠B and ∠FEQ = ∠A.(6), D1, A5
(9)P, E, Q are collinear, so ∠PEH + ∠HEF + ∠FEQ = 180°(2), A9
(10)∠A + ∠B = 90° and 180° = 90° + 90°, so ∠HEF = 90°(8), (9), L1, A19, A20, A21, convention
(11)The same argument at F, G, H shows all four angles of quadrilateral EFGH are right angles.(5), (9), (10), L1, A5, A9
(12)Four sides equal to c and four right angles, so EFGH is a square of side c.(7), (11), convention
(13)The square PQRS is divided without overlap into the four triangles and EFGH.(2), (12), convention
(14)The four triangle areas sum to ab/2 taken four times, that is 2ab.(6), A8, A6, A18
(15)(a+b)² = 2ab + c²(12), (13), (14), A6, A7
∎
Lemma L3
Placing the same four triangles in the same square in Arrangement II gives (a+b)² = 2ab + a² + b²
Proof.
(1)Take a square PQRS of side a + b again.A3
(2)On side PQ take a point U with PU = a, and on side PS a point W with PW = a.(1), convention
(3)UQ = b and WS = b(1), (2), A15
(4)The line through U parallel to PS and the line through W parallel to PQ are each unique, and the two lines meet at a point Z.(2), A1
(5)The two lines divide the square into four pieces. Each piece is a quadrilateral with two pairs of opposite parallel sides and four right angles.(1), (4), convention
(6)Opposite sides have equal lengths, so the adjacent sides of the four pieces are (a, a), (b, a), (a, b), (b, b) respectively.(2), (3), (5), A4
(7)Hence the pieces are one square of side a, one square of side b, and two rectangles with adjacent sides a and b.(6), convention
(8)Drawing a diagonal in each rectangle divides it into two right triangles with legs a, b.(7), convention
(9)These four agree with triangle ABC in two sides and the included angle, so they are congruent to ABC. They are the same four used in Arrangement I.(8), D1, A5
(10)The four triangle areas sum to 2ab.(8), A8, A6, A18
(11)(a+b)² = a² + b² + 2ab(1), (7), (8), (10), A6, A7
(12)(a+b)² = 2ab + a² + b²(11), A19, A20
∎
Theorem T1
In the right triangle ABC, a² + b² = c² (proof by area)
Proof.
(1)(a+b)² = 2ab + c²L2
(2)(a+b)² = 2ab + (a² + b²)L3, A20
(3)2ab + c² = 2ab + (a² + b²)(1), (2), E
(4)c² + 2ab = (a² + b²) + 2ab(3), A19
(5)c² = a² + b², that is a² + b² = c²(4), A21, E
(6)The equation of Arrangement II differs from the expansion formula (a+b)² = a² + 2ab + b² only in the order of terms. The dissection of area and the algebraic expansion are the same thing.L3, A17, A19, A20
∎
Theorem T2
In the right triangle ABC, a² + b² = c² (a second proof, using no area)
Proof.
(1)∠A and ∠B are acute.L1
(2)Let H be the foot of the perpendicular from C to line AB; such an H exists uniquely.A13
(3)H is between A and B, so AH + HB = AB = c(1), (2), A14, A15, D1
(4)In triangles AHC and ACB, ∠A is common and ∠AHC = ∠ACB = 90°(2), D1
(5)Two angles are respectively equal, so triangles AHC and ACB are similar.(4), A11
(6)Corresponding side ratios are equal, so AH : AC = AC : AB(5), A12
(7)AC = b > 0, AB = c > 0, so AH × c = b²(6), A16, A22, D1
(8)In triangles BHC and BCA, ∠B is common and ∠BHC = ∠BCA = 90°(2), D1
(9)By the same argument the two triangles are similar and BH : BC = BC : BA(8), A11, A12
(10)BC = a > 0, BA = c > 0, so BH × c = a²(9), A16, A22, D1
(11)a² + b² = BH × c + AH × c = (BH + AH) × c(7), (10), A18, A19
(12)BH + AH = AH + HB = c(3), A19
(13)a² + b² = c × c = c²(11), (12), E, convention
∎
Remark
Can A1 be removed from these two proofs?
The two proofs use different material. One uses area, the other similarity. Check where the two routes meet.